Part CCSIR NET June 2024the-jump-condition-fixes-the-sign

The jump condition fixes the sign

Consider the boundary value problem (BVP) (e^(−5x)y′)′ + 6e^(−5x)y = −f(x), 0 < x < ln 2, with y(0) = 0, y(ln 2) = 0. If $G (x, ξ) = (e^{(} 3 x) +$ Be^(2x))(Ce $^{(} 2 ξ) +$ De $^{(} 3 ξ))$ for $0 \leq ξ \leq x$ , and $G (x, ξ) = (e^{(} 3 ξ) +$ Be $^{(} 2 ξ)) ($ Ce^(2x) + De^(3x)) for $x \leq ξ \leq ln 2 ($ Green's function) is such that $\int_{0}^{(} ln 2) G (x, ξ) f (ξ) d ξ$ is the solution of the BVP, then the values of B, C and D are

A.B = −2, C = −1, D = 1
B.B = −2, C = 1, D = −1✓
C.B = 2, C = 1, D = 1
D.B = 2, C = −1, D = −1

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward.

See pricing

50 are analysed free — try those first.

Related counterexample: Every boundary value problem has a Green's function

The jump condition fixes the sign

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